This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Orthogonal Bases and the QR Algorithm by Peter J. Olver University of Minnesota 1. Orthogonal Bases. Throughout, we work in the Euclidean vector space V = R n , the space of column vectors with n real entries. As inner product, we will only use the dot product v · w = v T w and corresponding Euclidean norm bardbl v bardbl = √ v · v . Two vectors v , w ∈ V are called orthogonal if their inner product vanishes: v · w = 0. In the case of vectors in Euclidean space, orthogonality under the dot product means that they meet at a right angle. A particularly important configuration is when V admits a basis consisting of mutually orthogonal elements. Definition 1.1. A basis u 1 ,..., u n of V is called orthogonal if u i · u j = 0 for all i negationslash = j . The basis is called orthonormal if, in addition, each vector has unit length: bardbl u i bardbl = 1, for all i = 1 ,...,n . The simplest example of an orthonormal basis is the standard basis e 1 = 1 . . . , e 2 = 1 . . . , ... e n = . . . 1 . Orthogonality follows because e i · e j = 0, for i negationslash = j , while bardbl e i bardbl = 1 implies normality. Since a basis cannot contain the zero vector, there is an easy way to convert an orthogonal basis to an orthonormal basis. Namely, we replace each basis vector with a unit vector pointing in the same direction. Lemma 1.2. If v 1 ,..., v n is an orthogonal basis of a vector space V , then the normalized vectors u i = v i / bardbl v i bardbl , i = 1 ,...,n , form an orthonormal basis. Example 1.3. The vectors v 1 = 1 2 − 1 , v 2 = 1 2 , v 3 = 5 − 2 1 , are easily seen to form a basis of R 3 . Moreover, they are mutually perpendicular, v 1 · v 2 = v 1 · v 3 = v 2 · v 3 = 0, and so form an orthogonal basis with respect to the standard dot 6/5/10 1 c circlecopyrt 2010 Peter J. Olver product on R 3 . When we divide each orthogonal basis vector by its length, the result is the orthonormal basis u 1 = 1 √ 6 1 2 − 1 = 1 √ 6 2 √ 6 − 1 √ 6 , u 2 = 1 √ 5 1 2 = 1 √ 5 2 √ 5 , u 3 = 1 √ 30 5 − 2 1 = 5 √ 30 − 2 √ 30 1 √ 30 , satisfying u 1 · u 2 = u 1 · u 3 = u 2 · u 3 = 0 and bardbl u 1 bardbl = bardbl u 2 bardbl = bardbl u 3 bardbl = 1. The appearance of square roots in the elements of an orthonormal basis is fairly typical. A useful observation is that any orthogonal collection of nonzero vectors is automati cally linearly independent....
View
Full
Document
This note was uploaded on 02/10/2012 for the course MATH 5485 taught by Professor Olver during the Fall '09 term at University of Central Florida.
 Fall '09
 Olver
 Vectors, Vector Space

Click to edit the document details